Analog10y

PROBLEM 053 · Expert · The current-mirror OTA

Twice as Nice

A current-mirror OTA with a gain of two

The problem

Build a current-mirror OTA. An NMOS pair (gates “inp” and “inn”, tail Current source) has a diode-connected PMOS load on each drain (M3 on the “inp” side, M4 on the “inn” side). Each diode is mirrored by a PMOS with twice its width: M5 copies M3 and feeds “out” directly; M6 copies M4 and feeds a diode-connected NMOS (M7), which an NMOS (M8) of the same size mirrors to pull current from “out”. The output is the junction of M5’s and M8’s drains.

The output current is K·(I3 − I4) with K = 2: the effective transconductance is K·gm1, twice that of the five-transistor OTA from the same pair. With the output resistance of only two transistors (ro5 ∥ ro8) the gain is K·gm1·Rout, and the bandwidth K·gm1/(2π·C_L) is twice as high. The price is two more mirrors, which add poles of their own.

The test bench

Numbers to hit

Design it yourself

Draw the circuit on a schematic canvas, size every transistor, and see it simulated in your browser with ngspice and the SkyWater SKY130 models. Every transistor lights up green, gold or red by operating region as you move the sliders.

Sign in or create a free account →

More on operational amplifiers